In my previous post, I wrote the question incorrectly. Please help!! $\displaystyle 2^ {x^2} \;=\;32(2^{4x}) $
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Originally Posted by needhelpplease1 In my previous post, I wrote the question incorrectly. Please help!! $\displaystyle 2^ {x^2} \;=\;32(2^{4x}) $ $\displaystyle 2^ {x^2} \;=\;32(2^{4x})\Leftrightarrow 2^{x^2}=2^{5+4x}\implies x^2=5+4x$
Originally Posted by felper $\displaystyle 2^ {x^2} \;=\;32(2^{4x})\Leftrightarrow 2^{x^2}=2^{5+4x}\implies x^2=5+4x$ thanks, but how do you solve for x?
Originally Posted by needhelpplease1 thanks, but how do you solve for x? The solution that you obtain in that equation, is the same of your original equation!
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